剑指Offer II 018.有效的回文

题目：

暴力思路：代码： 优化思路：

代码：

判断该字符是否合法，然后取出s的值，用StringBuffer接收s的值进行reverse和原数比较

package LeetCode.MarchSoft.SwordTowards.Day3Sword;
public class ZZ18 {
    public static void main(String[] args) {
    String s = "OP";
        System.out.println(isPalindrome(s));
    }
    public static boolean isPalindrome(String s) {

        String ss =s.toLowerCase();
        int countShu = 0;  //测量ss的有效长度
        int count = 0;
        for (int i = 0; i ='a'&&ss.charAt(i)<='z'||ss.charAt(i)>='0'&&ss.charAt(i)<='9'){
                countShu++;
            }
        }
        char[] nums = new char[countShu];  //这里使用countShu给数组赋值
        for (int i = 0; i ='a'&&ss.charAt(i)<='z'||ss.charAt(i)>='0'&&ss.charAt(i)<='9'){
                nums[count] = ss.charAt(i);
                count++;
            }
        }
        String s1 = String.valueOf(nums);  //取值
        int left =0,right = nums.length-1;
        boolean flag  = true;
        StringBuffer sb = new StringBuffer(s1).reverse();  //转换
        for (int i = 0; i  
优化思路： 
一样的思路，只不过代码简单了，但是用了char的isLetterOrDigit方法
 
代码： 
class Solution {
    public static boolean isPalindrome(String s) {
             StringBuilder str = new StringBuilder();
        for (int i = 0; i